What would be the best(simple) way if possible to drop 5vac 3amp to 2.5v at 2.5a for heaters for a single 2a3 tube??? Sorry for the dumb ? Hope someone has a simple answer. Thanks in advance.
An additional transformer or altered existing winding.
All other solutions will waste power or potentially cause noise.
All other solutions will waste power or potentially cause noise.
Arnulf is right.
However, you could rectify it, and put a 2.5A CCS behind it. Causes a lot of heat in the ccs though...
However, you could rectify it, and put a 2.5A CCS behind it. Causes a lot of heat in the ccs though...
Ohms law -> V = IR
R = V/I
2.5/2.5 = 1 ohm
The heater has a resistance of 1 ohm with 2.5 V across it and 2.5 amps of current going through it. A dropping resistor will have the same current going through it and must have a voltage across it of 2.5 volts (2.5 + 2.5 = 5).
Use a 1 ohm dropping resistor.
Also see Voltage divider - Wikipedia, the free encyclopedia
R = V/I
2.5/2.5 = 1 ohm
The heater has a resistance of 1 ohm with 2.5 V across it and 2.5 amps of current going through it. A dropping resistor will have the same current going through it and must have a voltage across it of 2.5 volts (2.5 + 2.5 = 5).
Use a 1 ohm dropping resistor.
Also see Voltage divider - Wikipedia, the free encyclopedia
Could you drop with an RC filter and eliminate some of the noise the other guys are commenting on?
If "R" part is doing the dropping it will heat up (inclusion of a capacitor doesn't affect that), as mentioned above. An alternative would be capacitive dropper, 2600 uF in series with the filament for 1R @ 60 Hz (mains frequency in the US). The problem is finding 2600 uF non-polarized capacitor, but one could substitute two larger capacitors in series (inverse) instead.
This might actually be the best solution as - just like transformer - it doesn't waste power in the form of heat, and on top of everything it also doesn't cost nearly as much or weigh as much as a suitable transformer.
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